2006 AIME A Problems/Problem 14
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Problem
A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground In setting up the tripod, the lower 1 foot of one leg breaks off. Let be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then can be written in the form where and are positive integers and is not divisible by the square of any prime. Find (The notation denotes the greatest integer that is less than or equal to )
Solution
From what we are given it easily follows that is an equilateral triangle, and its center is at the same time the foot of the height from . We know that and , from the Pythagorean theorem it follows that . The segment is of the segment . Then has length . is the height in the triangle , hence we have , and therefore . We can now compute the area of the triangle as , and finally the volume of the tetrahedron as .
Why did we compute all this stuff? Consider what happens when the leg breaks in the described place :
import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label("$A$", A, W); label("$B$", B, E); label("$E$", BB, E); label("$C$", C, W); label("$O$", O, SE); label("$D$", D, Z); label("$B'$", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)')
The triangle will now be the base of the tripod, and our goal is to compute the height from onto the plane .
First, consider the tetrahedron . It has the same base as , and as is in of , from similarity it follows that its height is , and therefore the volume of is of the volume of .
But this means that we know the volume of . All we have to do is to find the area of , then we can easily compute the answer we need.
Set up a coordinate system where , , and . Here we have , and .
Hence .
We can now compute the area of as .
And finally, from the fact that the volume of is of the volume of we have:
And as the prime factorization of is , we can conclude that and .
As and , we have , and hence .
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |